In the 2nd question of class test of Solutions of Triangles: II(JEE Live), please give me the detailed simplification of the 1st,2nd and 3rd steps whose picture is attached below

Dear Student,
Please find below the solution to the asked query:

As tanθ=sinθcosθtanA-B2=sinA-B2cosA-B2tanA-B2=sin2A-B2cos2A-B2....iNowcos2A=1-2sin2A or cos2A=2cos2A-1sin2A=1-cos2A2 or cos2A=1+cos2A2Hence i becomestanA-B2=sin2A-B2cos2A-B2=1-cos2A-B21+cos2A-B2=1-cosA-B1+cosA-B

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